具有身份验证的urllib.request.urlopen（url）

Question

我一直在玩漂亮的汤和解析网页几天。在编写的所有脚本中，我一直使用一行代码作为救星。代码行是：

r = requests.get('some_url', auth=('my_username', 'my_password')).

但是…

我想对（OPEN A URL WITH AUTHENTICATION）做同样的事情：

(1) sauce = urllib.request.urlopen(url).read() (1)
(2) soup = bs.BeautifulSoup(sauce,"html.parser") (2)

我无法打开网址并阅读需要身份验证的网页。我如何实现这样的目标：

  (3) sauce = urllib.request.urlopen(url, auth=(username, password)).read() (3) 
instead of (1)

Answer 1

看看官方文档中的如何使用urllib软件包获取Internet资源：

# create a password manager
password_mgr = urllib.request.HTTPPasswordMgrWithDefaultRealm()

# Add the username and password.
# If we knew the realm, we could use it instead of None.
top_level_url = "http://example.com/foo/"
password_mgr.add_password(None, top_level_url, username, password)

handler = urllib.request.HTTPBasicAuthHandler(password_mgr)

# create "opener" (OpenerDirector instance)
opener = urllib.request.build_opener(handler)

# use the opener to fetch a URL
opener.open(a_url)

# Install the opener.
# Now all calls to urllib.request.urlopen use our opener.
urllib.request.install_opener(opener)